At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to 7.5 N/kg?

Question

Answer given: 915.8 km

Charlotte Aaron · Accepted Answer

This occurs at a distance of #7.3xx10^6# m from the centre of the Earth (which is 900 km from the surface).

Is it acceptable for us to begin with the gravitational field field equation for Earth? #g=(GM)/r^2# where M is the mass of the Earth. You now modify the equation's subject to make it solve for r: #r = sqrt((GM)/g)# Once you enter the numbers, you're done! #r = sqrt(((6.67xx10^(-11))(5.98xx10^(24)))/7.5)# = #7.3 xx 10^6# m (By the way, since this value is from the centre of the Earth, and the radius of the Earth is #6.4 xx 10^6# m, we are looking at a point only 900 000 m (900 km) above the surface of the Earth.)

Isabella Alvarez · Answer

See below.

Considering #G = 6.673 10^(-11)# #M_(earth) = 5.98 10^24# #R_(earth) = 6.38 10^6# All in SI units. We need the value of #r# such that #(G M_(earth))/(R_(earth)+r)^2=7.5# Solving for #r# we get #r = 914.2# Km

Lily Adams · Answer

undefined The height above the Earth's surface where the gravitational field strength equals 7.5 N/kg can be found using the formula:

$$ g' = \dfrac{g}{(1 + \dfrac{h}{R})^2} $$

where:
- $ g' $ is the gravitational field strength at height $ h $ above the Earth's surface,
- $ g $ is the gravitational field strength at the Earth's surface (approximately 9.8 N/kg),
- $ h $ is the height above the Earth's surface,
- $ R $ is the radius of the Earth (approximately 6,371,000 meters).

Rearrange the formula to solve for $ h $:

$$ h = R \times \left( \left(\dfrac{g}{g'}\right)^{1/2} - 1 \right) $$

Substitute the given values to find $ h $.